Question

Consider the contour integral \( \oint_C \frac{dz}{z^4 + z^3 - 2z^2} \), along the curve \( |z| = 3 \) oriented in the counterclockwise direction. If Res[f, z\(_0\)] denotes the residue of f(z) at the point z\(_0\), then which of the following are TRUE?

• A. Res[f, 0] = -1/4
• B. Res[f, 1] = 1/3
• C. Res[f, -2] = -1/12
• D. Res[f, 2] = -1

Hint:

For calculating residues at simple poles where the function is a ratio \(P(z)/Q(z)\), a useful shortcut is \( \text{Res}[f, z_0] = \frac{P(z_0)}{Q'(z_0)} \). For example, at z=1, \(P(z)=1\), \(Q'(z) = 4z^3+3z^2-4z\), so \(Q'(1) = 4+3-4=3\). The residue is \(1/3\). This is often faster than the limit definition.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem requires finding the residues of a complex function at its singular points. The Residue Theorem is then used to evaluate the contour integral, but the question only asks to verify the values of the residues themselves. The first step is to find the poles of the function and their order.
Step 2: Key Formula or Approach:
The function is \( f(z) = \frac{1}{z^4 + z^3 - 2z^2} \).
1. Find the poles by setting the denominator to zero.
2. Factor the denominator to identify the poles and their orders.
3. Calculate the residue at each pole using the appropriate formula:
- For a simple pole at \(z_0\): \( \text{Res}[f, z_0] = \lim_{z \to z_0} (z-z_0)f(z) \).
- For a pole of order \(m\) at \(z_0\): \( \text{Res}[f, z_0] = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} [(z-z_0)^m f(z)] \).
Step 3: Detailed Calculation:
1. Find the poles:
\[ z^4 + z^3 - 2z^2 = z^2(z^2 + z - 2) = z^2(z+2)(z-1) = 0 \] The poles are:
- \( z = 0 \) (a pole of order m=2)
- \( z = 1 \) (a simple pole, m=1)
- \( z = -2 \) (a simple pole, m=1)
The curve is \( |z|=3 \), a circle of radius 3 centered at the origin. All three poles (0, 1, -2) are inside this curve.
2. Calculate the residues:
- (A) Residue at \(z=0\) (pole of order 2):
\[ \text{Res}[f, 0] = \frac{1}{(2-1)!} \lim_{z \to 0} \frac{d}{dz} [z^2 f(z)] = \lim_{z \to 0} \frac{d}{dz} \left[ \frac{1}{(z+2)(z-1)} \right] \] \[ = \lim_{z \to 0} \frac{d}{dz} \left[ (z^2+z-2)^{-1} \right] = \lim_{z \to 0} \left[ -1(z^2+z-2)^{-2}(2z+1) \right] \] \[ = \lim_{z \to 0} \frac{-(2z+1)}{(z^2+z-2)^2} = \frac{-(1)}{(-2)^2} = -\frac{1}{4} \] Statement (A) is TRUE.
- (B) Residue at \(z=1\) (simple pole):
\[ \text{Res}[f, 1] = \lim_{z \to 1} (z-1)f(z) = \lim_{z \to 1} \frac{1}{z^2(z+2)} = \frac{1}{1^2(1+2)} = \frac{1}{3} \] Statement (B) is TRUE.
- (C) Residue at \(z=-2\) (simple pole):
\[ \text{Res}[f, -2] = \lim_{z \to -2} (z+2)f(z) = \lim_{z \to -2} \frac{1}{z^2(z-1)} = \frac{1}{(-2)^2(-2-1)} = \frac{1}{4(-3)} = -\frac{1}{12} \] Statement (C) is TRUE.
- (D) Residue at \(z=2\):
\(z=2\) is not a pole of the function. Therefore, the residue at this point is not defined in this context, or is zero. The statement is irrelevant/false.
Step 4: Why This is Correct:
The calculations for the residues at the three poles z=0, z=1, and z=-2 are performed correctly using the standard formulas for poles of order m. All three statements (A), (B), and (C) yield the correct values.